Included in this hand-out are five examples of problems requiring the solution of a system of linear algebraic equations.

Transcription

1 he Lecture Notes, Dept. of heical Engineering, Univ. of TN, Knoville - D. Keffer, 5/9/98 (updated /) Eaple pplications of systes of linear equations Included in this hand-out are five eaples of probles requiring the solution of a syste of linear algebraic equations. The eaples are:. Steady state ass balances on a single-stage liquid-liquid etractor B. Steady state ass balances on a flash tan 4. heical Reaction Equilibria 7 D. Deterination of an independent set of cheical reactions E. Noral Mode nalysis of the Vibrational Spectru of a Molecule. Steady state ass balances on a single-stage liquid-liquid etractor onsider an etractor: Etract E, { E,b, E,c, E,f } etractor Solvent S, { S,b, S,c, S,f } Feed F, { F,b, F,c, F,f } Raffinate R, { R,b, R,c, R,f } This unit reoves uses a recycled furfural strea as the solvent to etract benzene fro a cycloheane product strea. The data you are given is F ol F,b. F,c.9 F,f. / hr S 5 ol / hr S,b. S,c. S,f.9989 R 95 ol / hr R,b R,c R,f??? E 55 ol / hr E,b E,c E,f??? E,b The equilibriu constants are: Kb. R,b E,c and Kc. 5. R,c You have si unnons, the copositions of the raffinate strea and the coposition of the etract strea. (a) Write si independent equations benzene ole balance: cycloheane ole balance: furfural ole balance: FF,b + SS,b RR,b E FF,c + SS,c RR,c E F + S R E E, b E, c F,f S,f R,f E, f (not used, dependent)

2 he Lecture Notes, Dept. of heical Engineering, Univ. of TN, Knoville - D. Keffer, 5/9/98 (updated /) raffinate ole fraction constraint: etract ole fraction constraint: R,b + R,c E,b E,c R, f E, f E,b benzene equilibriu constraint: Kb. R,b E,c c-heane equilibriu constraint: Kc. 5 R,c (b) Put equations in linear for benzene ole balance: R R,b + EE,b FF,b + S R R,c + EE,c FF,c + S R R,f EE,f FF,f + S R,b + R,c + R, f + + cycloheane ole balance: S, c furfural ole balance: + S, f (not used, dependent) raffinate ole fraction constraint: etract ole fraction constraint: E,b E,c E, f benzene equilibriu constraint: E,b R,bKb c-heane equilibriu constraint: E,c R,cK c (c ) Put equations in atri for atri of coefficients, (6 6) eqn/var R,b R, c R, f E, b E, c R E R E 4 5 K b 6 K c vector of right hand sides, b (6) eqn b F F,b + S F + S F,c (d) opute the deterinant and ran of the atri. det.89e+5 ran() 6 S,b S,c E, f S, b

3 he Lecture Notes, Dept. of heical Engineering, Univ. of TN, Knoville - D. Keffer, 5/9/98 (updated /) ran( b) 6 (e) Using MTLB, solve for the steady-state values of the unnons. (). R, b ().8448 R, c ().59 R, f (4).65 E, b (5).4 E, c (6).897 E, f

4 he Lecture Notes, Dept. of heical Engineering, Univ. of TN, Knoville - D. Keffer, 5/9/98 (updated /) B. Steady state ass balances on a flash tan onsider an isotheral flash tan: F, {z} Isotheral Flash Tan V, {y} L, {} This unit taes a pressurized liquid, three-coponent feed strea and eposes it to a lo pressure vessel aintained under isotheral conditions. The net result is that soe of the fluid is vaporized, hile soe fluid reains liquid. The copositions of the liquid and vapor phase are deterined by the cobined analysis of ass balances and Raoult s La for vapor-liquid equilibriu. The teperature in the flash tan is T 98K and the pressure in the tan is P Pa. Raoult s La states that the product of the liquid ole fraction of coponent i and the vapor pressure of coponent i is equal to the partial pressure of coponent i in the vapor phase: vap i i P y P i Use the folloing data for the teperature given above P T P vap T P T 98K 98K 98K F ol /hr z z z B.4.. V 44.78ol /hr y y y B??? L F - V ol /hr B??? Then you have si unnons, the copositions of the liquid strea and the coposition of the vapor strea. (a) Write equations ole balance: B ole balance: ole balance: liquid ole fraction constraint: vapor ole fraction constraint: Fz L Vy FzB LB VyB (not used, dependent) Fz L Vy (not used, dependent) + B + y + y + y B 4

5 he Lecture Notes, Dept. of heical Engineering, Univ. of TN, Knoville - D. Keffer, 5/9/98 (updated /) equilibriu constraint: B equilibriu constraint: equilibriu constraint: (b) Put equations in linear for ole balance: B ole balance: ole balance: vap vap B PB vap P P y y y B P P P L + Vy Fz L L B + VyB FzB (not used, dependent) + Vy Fz (not used, dependent) liquid ole fraction constraint: + B + vapor ole fraction constraint: y + yb + y equilibriu constraint: vap P y P B equilibriu constraint: vap B PB ybp equilibriu constraint: vap P yp (c ) Put equations in atri for atri of coefficients, (6 6) eqn/var B L V 4 vap P P 5 vap P B 6 vap P vector of right hand sides, b (6) eqn b y P Fz y B y P (d) opute the deterinant and ran of the atri. (Here is the contents of a Matlab -file I used to do this.) F ; V 44.78; 5

6 he Lecture Notes, Dept. of heical Engineering, Univ. of TN, Knoville - D. Keffer, 5/9/98 (updated /) L F - V; z.4; zb.; z.; Pvap.6; PvapB.; Pvap.; P.5; [L V Pvap -P PvapB -P Pvap -P] b [F*z; ; ; ; ; ] ran ran() det det() \b (); B (); (); y (4); yb (5); y (6); The output fro the code yielded: ran 6 det [ ].489 y.897 B.8 y B y.45 6

7 he Lecture Notes, Dept. of heical Engineering, Univ. of TN, Knoville - D. Keffer, 5/9/98 (updated /). heical Reaction Equilibria onsider that you have a three-coponent reactive iture, all undergoing reversible reactions, as pictured belo: In this picture, the s are concentrations of the three species and the s are rate constants. n eaple of this syste is the inetic equilibriu beteen para-, eta-, and ortho-ylene. No suppose e ant to no hat the concentration is as a function of tie. We can rite the ass balances for each coponent. There are no in and out ters (the reactor is a batch reactor). There is only the accuulation ter and the reaction ters. lso, assue each reaction is first order in concentration. d dt d dt d dt (8.6) We can gather lie ters and rearrange the right hand side: d dt d dt d dt ( + ( + ) + + ( ) ) (8.7) and e change this syste of equations into atri & vector for: d dt X (8.8) 7

8 he Lecture Notes, Dept. of heical Engineering, Univ. of TN, Knoville - D. Keffer, 5/9/98 (updated /) here X ( + ) ( + ) ( + ) (8.9) ne underbar denotes a colun vector; to underbars denotes a atri. This is a syste of linear differential equations. If e ant the steady-state solution to the differential equations, e set the accuulation ter to zero. Then e have a syste of linear algebraic equations, as shon: X (8.) Let s solve for the steady-state concentration. The rate constants are given as:.5 sec.5 sec. sec.5 sec. sec.5 sec The deterinant of X is and the ran is. Therefore e have an infinite nuber of solutions. Why? If e loo at X, e see that RW -RW - RW. Since equation is not linearly independent e can drop it. Then e have equation and three unnons. We have options. ption Nuber ne: Since, e have an infinite nuber of solutions, e can just ae one of the variables no longer variable. Then e is no longer a variable. ill have the rest of relative to this basis. This requires us to rearrange X since We coe up ith the ne equations here X b X ( + ) ( + ) b 8

9 he Lecture Notes, Dept. of heical Engineering, Univ. of TN, Knoville - D. Keffer, 5/9/98 (updated /) ne can see that e have dropped the third equation and since is no longer a variable, e have oved it to the right hand side of the equations. (If you don t see this, right these equations out in non-atri for and then ove to the right-hand-side of the equation, and then change bac into atri for. You ill get this result.) With these ne definitions, e can pic a value for MTLB gives the inv() upon request and provides the solution, lie. No MTLB says det() hich gives us the eaple solution vector:.5.5 If e ant to find the olar copositions, then e ill require that i. In order to noralize our i solution vector so that they su to one, e use the standard noralization equation: z i i i i then e find the steady state olar copositions to be: z / 7 / 7 4 / 7. ption Nuber To: If e only have equations for unnons, e can find another independent equation. If our unnons are ole fractions e no they ust su to one. i i This is our third equation. We drop equation three fro the previous forulation of the proble and e have X b 9

10 he Lecture Notes, Dept. of heical Engineering, Univ. of TN, Knoville - D. Keffer, 5/9/98 (updated /) here X ( + ) ( + ) b This atri has a deterinant of.455 and ran of. Therefore e have a set of independent equations. Solving for yields / 7 / 7 4 / 7 hich is the sae result e obtained doing the proble the other ay.

11 he Lecture Notes, Dept. of heical Engineering, Univ. of TN, Knoville - D. Keffer, 5/9/98 (updated /) D. Deterination of an independent set of cheical reactions Proble. Deterine the nuber of independent reaction in the folloing set by eaining the ran of the stoichioetric coefficient atri. (In the stoichioetric coefficient, the ros represent reactions and the coluns represent olecules.) 4 NH N + 6 H 4 NH + 4 N + 6 H N + N 4 NH + 6 N 5 N + 6 H N + N H N N N NH reactions olecules This is a non-square atri, five ros, si coluns. We can do the sae NGE on a non-square atri as on a square atri to reach Upper triangular for STEP NE. (zero the first colun belo ro ) RW RW - RW RW4 RW4 - RW H N N 5 4 N NH reactions olecules STEP TW. (zero the second colun belo ro ) RW RW + ½*RW RW4 RW4-5/*RW RW5 RW5 - ½*RW

12 he Lecture Notes, Dept. of heical Engineering, Univ. of TN, Knoville - D. Keffer, 5/9/98 (updated /) reactions NH 4 5 olecules N N N H 6 There are only three independent cheical reactions. This ind of analysis is iportant hen you are doing an process analysis on a reactive syste. You need to have only independent reactions incorporated in your aterial and energy balances. The stoichioetric coefficient atri tells the engineer, given a set of reactions, ho any of those reactions are (linearly) independent. It ill not give the engineer a coplete set of reactions, unless () the nuber of reactions in the set is greater than or equal to the total nuber of independent reactions and () all coponents in the syste appear in at least one reaction. Proble. Deterine the nuber of independent reactions by using the atoic atri. (The atoic atri has atos on the ros and olecules along the coluns atos N N H olecules NH N N H This atri is already in upper triangular for. The ran of the atri is three. In reactive process analysis, the total nuber of independent reactions is equal to the nuber of coponents less the ran of ato atri. Since e have si coponents and a ran of three, e have three independent reactions. The atoic atri tells the engineer ho any independent reactions there can be. This is different in several ays fro the stoichioetric atri above. Fro atoic atri, # of independent reactions # of cheical species - ran of atoic atri If you ant to find a set of three reactions that are coplete and independent, then you ust postulate three reactions, and use the stoichioetric atri technique, outlined above, to verify that they are independent. The reactions you postulate ust contain all of the cheical species. Here is an eaple set of reactions: N + N N + N 4 NH N + 6 H These are guaranteed to be linearly independent since each reaction contains a unique species. Reaction is the only reaction to contain N. Reaction is the only reaction to contain N. Reaction is the only reaction to contain NH.

13 he Lecture Notes, Dept. of heical Engineering, Univ. of TN, Knoville - D. Keffer, 5/9/98 (updated /) E. Noral Mode nalysis of the Vibrational Spectru of a Molecule onsider that e ant to investigate the vibrational properties of carbon dioide,. ur odel of the olecule loos lie this: spring () spring (),,, energy, U, is We odel the interaction beteen olecules as Hooian springs. For a Hooian spring, the potential U ( ) (8.) and the force, F, is F ( ) (8.) here is the spring constant (units of g/s ), is the equilibriu displaceent, and is the actual displaceent. We can rite Neton s equations of otion for the three olecules: a F a F a F ( ) ( ) + ( ) ( ) (8.) Knoing that the acceleration is the second derivative of the position, e can rerite the above equations in atri for as (first divide both side of all of the equations by the asses) d dt (8.4) here

14 he Lecture Notes, Dept. of heical Engineering, Univ. of TN, Knoville - D. Keffer, 5/9/98 (updated /) 4 (8.5) Solving this syste of second order linear differential equations yields the integrated equations of otion for carbon dioide. Hoever, even ithout solving this syste, because the syste is linear, there are special techniques e can use to tell us a lot about the syste s behavior. Let us deterine the eigenvalues and the eigenvectors of the atri. First e have: λ λ λ λ I (8.6.5) The characteristic equation is given by (using equation 8.9) ( ) I det λ λ λ λ λ λ (8.6.6) Solving this for λ, e have ( ) I det λ + + λ λ λ λ (8.6.7) ( ) I det + λ λ λ λ (8.6.8) ( ) I det + λ + λ + λ λ (8.6.9)

15 he Lecture Notes, Dept. of heical Engineering, Univ. of TN, Knoville - D. Keffer, 5/9/98 (updated /) 5 So the roots to the characteristic equation are + λ λ λ (8.6.) The eigenvectors for each of these eigenvalues are given by ( ) I λ (8.6.4) ( ) I λ (8.6.) since the deterinant of is zero, e cannot use the inverse to calculate the solution, but e can still use NGE if e augent the atri by the solution vector. Since the equations are not linearly independent, e can reove, as a variable and set it equal to one. Then our syste becoes:,, (8.6.) With this ne atri, e can calculate the deterinant ) det( so it has an inverse hich is So that the solution to equation (8.6.) is,,

16 he Lecture Notes, Dept. of heical Engineering, Univ. of TN, Knoville - D. Keffer, 5/9/98 (updated /) 6 and the eigenvector that corresponds to λ is (8.6.5) To find the second eigenvector, the eigenvector that corresponds to λ λ I + λ I I) det( λ s before, e reove the third equation and the third variable fro the equation: +,, (8.6.6) With this ne atri, e can calculate the deterinant ) det( so it has an inverse hich is + So that the solution to equation (8.6.6) is

17 he Lecture Notes, Dept. of heical Engineering, Univ. of TN, Knoville - D. Keffer, 5/9/98 (updated /) 7,, and the eigenvector that corresponds to λ is (8.6.7) To find the third eigenvector, the eigenvector that corresponds to + λ λ I I) det( λ s before, e reove the third equation and the third variable fro the equation:,, (8.6.8) With this ne atri, e can calculate the deterinant ) det( so it has an inverse hich is So that the solution to equation (8.6.6) is

18 he Lecture Notes, Dept. of heical Engineering, Univ. of TN, Knoville - D. Keffer, 5/9/98 (updated /) 8,, and the eigenvector that corresponds to + λ is (8.6.9) So e have the three eigenvalues and the three eigenvectors. So hat? What good do they do us? For a vibrating olecule, the square root of the absolute value of the eigenvalues fro doing an eigenanalysis of Neton s equations of otion, as e have done, are the noral frequencies. You see that the units of the eigenvalues are /sec, so the square root has units of frequency (or inverse tie). For carbon dioide, the three noral frequencies are: + ω The frequency of zero is no frequency at all. It is not a vibrational ode. In fact, it is a translation of the olecule. We can see this by eaining the eigenvectors. The eigenvector that corresponds to λ or λ ω is (8.6.5) This is a description of the noral vibration associated ith eigenfrequency of zero. It says that all atos ove the sae aount in the -direction.

19 he Lecture Notes, Dept. of heical Engineering, Univ. of TN, Knoville - D. Keffer, 5/9/98 (updated /) The eigenvector that corresponds to λ or ω λ is (8.6.7) This eigenvector describes a vibration here both the oygen ove aay fro the equally and the carbon does not ove. The eigenvector that corresponds to λ + or ω λ + is (8.6.9) This eigenvector describes a vibration here both the oygen ove to the right and the carbon ove ore to the left, in such a ay that there is no center of ass otion. 9

20 he Lecture Notes, Dept. of heical Engineering, Univ. of TN, Knoville - D. Keffer, 5/9/98 (updated /) The noral odes of otion provide a coplete, independent set of vibrations fro hich any other vibration is a linear cobination.